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Chapter 41: Q. 8 (page 1207)

Predict the ground-state electron configurations of $M g, S r$ and $B a$ ?

Short Answer

Expert verified

The ground-state electron configurations of $M g$ is $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}$ , $S r$ is $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{2}$ and $B a$ is $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{2} 4 d^{10} 5 p^{6} 6 s^{2}$ .

Step by step solution

01

Given Information

We need to find the ground-state electron configurations of $M g, S r$ and $B a$ .

02

Simplify

The electron configurations for $M g, S r$ and $B a$ .

First, finding electron configuration for $M g (Z = 12)$ :

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}$

Total electrons is $(2 + 2 + 6 + 2 = 12)$ .

Now, finding electron configuration for role="math" localid="1650270308995" $S r (Z = 38)$ :

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{2}$

Total electrons is $(2 + 2 + 6 + 2 + 6 + 2 + 10 + 6 + 2 = 38)$

At last , finding electron configuration for $B a (Z = 56)$ .

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{2} 4 d^{10} 5 p^{6} 6 s^{2}$

Total electrons is $(2 + 2 + 6 + 2 + 6 + 2 + 10 + 6 + 2 + 10 + 6 + 2 = 56)$

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Most popular questions from this chapter

What is the electron configuration of the second excited state of lithium?

Predict the ground-state electron configurations of $A l, G n$ and $I n$ ?

Suppose you have a machine that gives you pieces of candy when you push a button. Eighty percent of the time, pushing the button gets you two pieces of candy. Twenty percent of the time, pushing the button yields $10$ pieces. The average number of pieces per push is $N_{a v g} = 2 \times 0.80 + 10 \times 0.20 = 3.6$ . That is, $10$ pushes should get you, on average, $36$ pieces. Mathematically, the average value when the probabilities differ is $N_{a v g} = \sum (N_{i} \times P r o b a b i l i t y o f i)$ . We can do the same thing in quantum mechanics, with the difference that the sum becomes an integral. If you measured the distance of the electron from the proton in many hydrogen atoms,

you would get many values, as indicated by the radial probability density. But the average value of $r$ would be

$r_{a v g 0} = \int_{0}^{\infty} r P_{r} (r) d t$

Calculate the average value of $r$ in terms of $a_{B}$ for the electron in the $1 s$ and the $2 p$ states of hydrogen.

In LASIK surgery, a laser is used to reshape the cornea of the eye to improve vision. The laser produces extremely short pulses of light, each containing $1.0 m J$ of energy.

a. There are $9.7 \times 10^{14}$ photons in each pulse. What is the wavelength of the laser?

b. Each pulse lasts a mere $20 n s$ . What is the average power delivered to the cornea during a pulse?

List the quantum numbers, excluding spin, of

(a) all possible $3 p$ states and

(b) all possible $3 d$ states.

See all solutions

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